MCAT Science Assessment- Organic Chemistry (W/Expln's)
What happens in aldol additions? (A) An enolate attacks a carbonyl carbon leading to a new carbon-carbon bond. (B) A carbonyl oxygen attacks a carbonyl carbon from another molecule. (C) An alcohol attacks another alcohol, loses water, and forms a new carbon-carbon bond. (D) An aldehyde and a carboxylic acid react to lose water and form a carbon-carbon bond.
(A) An enolate attacks a carbonyl carbon leading to a new carbon-carbon bond. // Aldol addition is a way of forming carbon-carbon bonds. It involves a nucleophilic attack on the carbonyl carbon of an aldehyde or ketone by an enolate ion.
What serves as the backbone for DNA? (A) Sugars linked by phosphodiester bonds to phosphate groups. (B) Sugars linked by sulfodiester bonds to sulfate groups. (C) Nitrogenous bases linked by phosphodiester bonds to phosphate groups. (D) Nitrogenous bases linked by sulfodiester bonds to sulfate groups.
(A) Sugars linked by phosphodiester bonds to phosphate groups. // DNA consists of a sugar-phosphate backbone that has nitrogenous bases attached to sugar rings. This means that the backbone only consists of sugars and phosphates.
(B) 5-Hydroxyhexanoic acid // The IUPAC naming conventions first involve the identification of the longest carbon chain with the highest priority group. In this case, the carboxylic group (-COOH) contains the longest carbon chain and the correct answer choice will end in "acid" like all carboxylic acids. The first carbon is attached to the carboxylic group and the hydroxyl group is attached to the 5th carbon.
(A) ß-Hydroxyhexanoic acid (B) 5-Hydroxyhexanoic acid (C) 6-carboxy-2hexanol (D) 1-carboxy-5-hexanol
Which of the following conditions could result in the formation of a double bond between carbon and nitrogen? (A) a hydrocarbon reacting with ammonia (B) A ketone reacting with ammonia (C) A carboxylic acid reacting with NO2 (D) An alcohol reacting with ammonia
(B) A ketone reacting with ammonia. // A carbon double bonded to nitrogen is called an imine and is made when an aldehyde or ketone is reacted with ammonia. The nitrogen displaces the oxygen originally double bonded to carbon and water is lost.
What is the product of reacting a ketone with LiAlH4? (A) A primary alcohol (B) A secondary alcohol (C) A carboxylic acid (D) An aldehyde
(B) A secondary alcohol // Lithium aluminum hydride, LiAlH4 , is a reducing agent and will force carbonyls to become alcohols. Since the question stem indicates the reduction of a ketone, the carbonyl carbon is already attached to two other carbons. Reduction doesn't break down C-C bonds and so these will stay intact, resulting in (B) , a secondary alcohol.
What is the first step in the Strecker synthesis of amino acids? (A) Nucleophilic attack on a nitrile carbon. (B) Aminonitrile formation. (C) Reaction of alkyl halide into a primary amine. (D) Electrophilic attack by ammonia.
(B) Aminonitrile formation // Strecker synthesis is a two-step process to form amino acids. The first step is aminonitrile formation followed by protonation of the aminonitrile and attack by water to turn the nitrile into a carboxylic acid.
AN IR peak in te region 1680-1750 cm(-1) is characteristic of what type of bond stretch? (A) O-H (B) C=O (C) C-C (D) C-H
(B) C=O //AN IR peak region of 1680-1750 cm-1 is representative of a carbonyl functional group. This is a high yield fact worth memorizing since carbonyl groups are found in a number of different organic compounds and ca be used to distinguish them from other commonly seen functional groups. Choice (A). a O-H group would correspond to a broad peak round 3000-3500 cm-1 Choice (C). a C-C group would correspond to a peak around 1200-1400 cm-1 Choice (D). a C-H group would correspond to a peak around 2800- 3000 cm-1
What is the order of increasing boiling points in the following compounds? I. CH3 CH2 COOH II. CH3 CH2 CHO III. CH3 CH2 CH2 OH (A) I < III < II (B) II < III < I (C) III < II < I (D) I < II < III
(B) II < III < I // Boiling points are a physical property of a compound that is determined soloed by the strength of the intermolecular forces. Intermolecular forces are the bonding interactions between two compounds unlike intramolecular forces that are within one compound. The greater the intermolecular forces between compounds, the greater the amount of energy required to separate these compounds in the solid or liquid phase so that they are liberated as a gas. This energy is usually supplied in the form of heat, so a compound with a high BP will require more energy. Identify the functional groups of each of the choices and determine the possible intermolecular forces that can result between two of the same compounds. Compound 1 is a carboxylic acid that can form two hydrogen bonds to another carboxylic acid. Compound 2 is an aldehyde that has a polar carbonyl group that can form dipole-dipole interactions. Compound 3 is a primary alcohol that can form one hydrogen bond to another primary alcohol. //H-bonds > dipole-dipole
Why might a scientist modify an alcohol into a tosylate before proceeding with his reaction? (A) To make the hydrogen on the alcohol more acidic (B) To turn the alcohol into a better leaving group (C) To make the solution more basic (D) To encourage more hydrogen bond to form
(B) To turn the alcohol into a better leaving group. // An alcohol is a poor leaving group and tosylates and mesylates help fix that. They displace OH— of the alcohol and make a new structure that is easily removable.
Which of the compounds below cannot be oxidized to a carboxylic acid? (A) an aldehyde (B) a tertiary alcohol (C) a primary alcohol (D) a terminal alkene
(B) a tertiary alcohol // Tertiary alcohols are not normally oxidized since a C-C bond would have to be broken to form a C-O bond. C-C bonds are very strong and stable, and thus resistant to an oxidizing agent. Choice (A). An aldehyde can easily be oxidized into a carboxylic acid. Aldehydes are one of the common reactants to produce carboxylic acids by oxidation. Choice (C). A primary alcohol can be easily oxidized into a carboxylic acid. Primary alcohol are the only alcohols that can be used to produce carboxylic acids by oxidation since oxidized secondary alcohols form ketones that cannot be oxidized further. Choice (D). A terminal alkene can be oxidized into a primary alcohol, so it is a possible substrate to form a carboxylic acid using an oxidizing agent.
The bond angle most associated with sp3 hybridized atoms is? (A) 104.5° (B) 120° (C) 109.5° (D) 180°
(C) 109.5° degrees A sp3 hybridized carbon has four single bonds attached to it so there is no electron repulsion of the bond angles. In the case of cyclohexane, these -C-C-C- bonds are sp3 hybridized and have angles of 109.5° Choice (A). This is the angle in sp3 hybridized atoms with two lone pairs causing repulsion of the bond angles to 104.5° . For instance, the O in water is sp3 hybridized, but has two lone pairs to cause bond angle repulsion. Choice (B) This is the angle in a sp2 hybridized atom with the planar geometry such as the atoms in a alkene. They will have bond angles of 120° . Choice (D) Bond angles of 180° are found in sp hybridized carbons that have triple bonds. These carbons will have a linear geometry, such as the carbons in alkynes.
Which of the following carboxylic acid derivatives are most reactive towards nucleophilic acyl substitution? (A) Esters (B) Acid anhydrides (C) Acyl halides (D) Amides
(C) Acyl halides // Carboxylic acid derivatives are acyl groups attached to halides (acyl halides), carboxyl group (acid anhydrides), amine (amides), or alkoxy groups (esters). The reactivity of a carboxylic acid derivative in nucleophilic acyl substitution is governed by the strength of the leaving group. Acyl halides are the most reactive carboxylic acid derivative and therefore the least stable, while amides are the least reactive but the most stable. Acid anhydrides are more reactive that esters, but less reactive than acyl halides. Therefore, the order of reactivity is acyl halides > acid anhydrides > esters > amides. Choice (A). Esters are one of the more stable carboxylic acid derivatives and therefore not as reactive as acyl halides or acid anhydrides. The leaving group for an ester is an alkoxy group (-OR), which is a poor leaving group. Choice (B). Acid anhydrides are the second most reactive carboxylic acid derivative, but not as strong as acyl halides. Choice (D). Amides are the least reactive and the most stable carboxylic acid derivative.
In chromatography, which of the following set of characteristics will allow a sample to travel the furthest? (A) High affinity to the mobile phase, high affinity to the stationary phase? (B) Low affinity to the mobile phase, high affinity to the stationary phase? (C) High affinity to the mobile phase, low affinity to the stationary phase? (D) Low affinity to the mobile phase, low affinity to the stationary phase?
(C) High affinity to the mobile phase, low affinity to the stationary phase. // For something to move in chromatography, it must have affinity for the mobile phase so it will move with it. For it to move far, it must have limited affinity for the stationary phase or else it will not travel well. This high affinity for mobile/low affinity for mobile/low affinity for stationary is characterized in choice (C).
When an alkene is reacted with ozone, ozonolysis occurs, wherein the double-bond is cleaved and both of the carbons involved in the double-bond become carbonyl carbons. What is the product of a tri-substituted alkene reacted with ozone, followed by hydrogen peroxide (an oxidixing agent)? (A) Two carboxylic acids (B) One carboxylic acid (C) One ketone and one carboxylic acid (D) Two aldehydes
(C) One ketone and one carboxylic acid // After ozonolysis the tri-substituted alkene will become a ketone and an aldehyde. Ketones cannot be oxidized, but aldehydes can be oxidized to carboxylic acids, so in the presence of an oxidizing agent, the expected products would be a ketone and a carboxylic acid, which matches (C).
How would the 1 H NMR change if ethane (CH3 CH3), were changed to chloroethane, CH3 CH2 Cl? (A) One singlet would become two doublets with one doublet shifted downfield. (B) One singlet would become a doublet and a triplet with the triplet shifted downfield. (C) One singlet would become a triplet and a quartet with the quartet shifted downfield. (D) Two quartets would become a triplet and a quartet with the quartet shifted downfield.
(C) One singlet would become a triplet and a quartet with the quartet shifted downfield. // In 1 H NMR spectroscopy, it is important to look at the different kinds of hydrogen present. In ethane, all of the hydrogen are identical which means that there will only be one singlet, a large peak, and since it is an alkane, it will not be very far downfield. In chloroethane, there are now two different types of hydrogen. The hydrogen closest to the chlorine should be deshielded by the electronegative character of chlorine and therefore be shifted downfield. The next thing to look at is splitting. Splitting follows the n+1 rule, where a type of hydrogen will be slit by how many hydrogens are on adjacent carbons. In chloroethane, the hydrogen proximal to the chlorine have three hydrogen on the adjacent carbon and thus will form a quartet. For the hydrogen distal to the chlorine, they have two hydrogen on an adjacent carbon and will therefore split into a triplet.
What aqueous reagent would be most suitable in the extraction of phenol from CCl4? (A) Sodium bicarbonate (NaHCO3) (B) Hydrochloric acid (HCl) (C) Sodium hydroxide (NaOH) (D) Sodium chloride (NaCl)
(C) Sodium hydroxide // The basic scheme behind an organic extraction is to isolate the organic molecule from the organic layer by moving it into the aqueous layer by protonation with an acid or deprotanation with a base. The aqueous layer can then be easily drained from the separator funnel and the organic molecule can be isolated. Phenol is a weakly acidic organic molecule, so only a strong base will deprotonate the molecule. If a weak base is used, it will not be sufficient to extract the proton from phenol, so it will remain in the organic CCl4 layer. Of the choices listed, only sodium hydroxide is a strong base. Choice (A). Sodium bicarbonate is a weak base and therefore will not deprotonate phenol so that it goes to the aqueous layer. Weak bases are used to deprotonate strong organic acids in extractions. Choice (B). Phenol is a weak acid so a strong acid like HCl would be ineffective in an extraction. Choice (D). Sodium chloride is neither an acid nor a base, so it would have no effect in an extraction.
What drives the first step of base-catalyzed keto-enol tautomerization? (A) The acidity of the proton on the alcohol. (B) The ketone oxygen being protonated. (C) The acidity of the proton alpha to the carbonyl. (D) Nucleophilic attack on the carbonyl carbon.
(C) The acidity of the proton alpha to the carbonyl. // The first step in base-catalyzed keto-enol tautomerization is the formation of a carbanion. This is accomplished by removing a proton from a carbon that is alpha to a carbonyl. Normally this proton is not removable, but when next to a carbonyl, the electronegativity of oxygen allows the proton to be removed and ketone-enol tautomerization to occur.
Which of the following cannot be oxidized to form a carboxylic acid? a) a primary alcohol b) an alkyl benzene c) a ketone d) an aldehyde
(C) a Ketone // One of the common ways to synthesize carboxylic acids is to use a strong oxidizing agent, such as PCC or potassium permanganate (KMnO4). Secondary or tertiary alcohols and ketones would be unsuitable as a starting reagent to produce a carboxylic acid because of valence limitations. Choice (A). A primary alcohol can be oxidized into a carboxylic acid since it terminates in a -OH group. Choice (B). An alkyl benzene can be converted to a carboxylic acid by oxidation. Choice (D). An aldehyde can be converted into a carboxylic acid by oxidation.
A compound with the general formula RCOOOCOR (where R= -H or any alkyl group) is classed as a(n): (A) ether (B) ester (C) acid anhydride (D) epoxide
(C) acid anhydride // The compound has two carbonyl oxygens with a shared oxygen between the two carbonyl carbons. That is the general form of an acid anhydride that results from the reaction of two carboxylic acids with the loss of water between them. Choice (A). An ether has no carbonyl groups, so this cannot be correct. The general form of an ether is (ROR') Choice (B). An ester has only one carbonyl group and results from the reaction of a carboxylic acid and an alcohol. Choice (D). An epoxide is a cyclic ether. They do not contain carbonyl groups.
Which of the following would produce a ketone when it undergoes Jones oxidation? (A) Ethanol (B) 1-propanol (C) isobutanol (D) 2-propanol
(D) 2-propanol // Jones oxidation, oxidizes primary alcohols to carboxylic acids and secondary alcohols into ketones. The only secondary alcohol present is choice (D), 2-propanol.
A scientist using UV-Vis spectroscopy would anticipate which of the following molecules to absorb the largest wavelength? (A) ethanol (B) methane (C) ammonia (D) benzene
(D) Benzene // UV-Vis spectroscopy measures the wavelengths that are absorbed by a molecule by shining light through a sample. Larger molecules can usually absorb larger wavelengths and conjugation allows even larger wavelengths to be absorbed. This means the molecule with the largest wavelength absorption should be the biggest one. Which matches to benzene, which also is conjugated allowing for even greater wavelength absorption.
Which of the following carboxylic acids will be the most acidic? (A) CCl3CH2COOH (B) CHF2CH2COOH (C) CH3CFClCOOH (D) CH3CF2COOH
(D) CH3CF2COOH // The most acidic organic molecules results from the greatest stabilization of the negative charge that results in the conjugate base when a proton is removed from the organic acid. Resonance or inductive forces can disperse this negative charge where halogen atoms are attached to the molecule to withdraw away the negative charge. Since all these answer choices are carboxylic acids and their only differences are the number and location of halogens to the carboxyl group the most acidic molecule will have the strongest inductive effects. Generally, fluorine has stronger inductive effects than chlorine and the electron withdrawing effects will be strongest when the halogens are attached closest to the negative charge. Therefore, (D) will be the strongest acid. Choice (A). This organic compound contains three chlorines, but the nature of halogen (F>Cl) and their proximity to negative charge have stronger effects in acidic trends. Choice (B). This organic compound contains two fluorines that are two carbons away from the negative charge. It will thus have a decreased effect in withdrawing negative charge from the conjugate base. Choice (C). This organic compound contains one chlorine and one fluorine that are one carbon away from the negative charge. It will thus have a decreased effect in withdrawing negative charge from the conjugate base compared to two fluorines in the same location.
Which group has the highest priority when naming a compounds? (A) Alcohol (B) Alkane term-26 (C) Aldehyde (D) Carboxylic Acid
(D) Carboxylic Acid // On the MCAT Carboxylic Acid is the highest priority group that will be tested. the carbon in R-COOH is the most oxidized because it has two bonds to Oxygen.
Which of these bonds would result in the shortest bond length in an atom? (A) One σ bond only (B) Two π bond only (C) Two σ bonds only (D) One σ bond and one π bond only
(D) One σ bond and one π bond only. // The first bond between any two molecules is a σ bond. There can only be one σ bond between any two molecules so this discounts (C) . After two molecules are bound with a σ bond, π bonds form to strengthen the bond. Without the σ bond, no π bonds can exist, which discounts (B) . Choice (D) is correct because it is the definition of a double bond which is shorted than a single bond.
Which of the following is NOT true about sp2 orbitals? a) The nitrogen in ammonia is sp2 hybridized b) Carbons double bonded to each other are sp2 hybridized c) They have 33% s character and 66% p character d) The sp2 orbitals are situated 120 apart
**(A)** // sp2 orbitals are orbitals that are a mixture of two p orbitals and one s orbital. This creates three orbitals that are 33% s and 66% p and are all situated 120° from each other. Common examples are alkenes, or carbon-carbon double bonds. The only one that doesn't fit here is (A) since nitrogen in ammonia is sp3 hybridized
(A) methoxybenzene; 1-methoxy-1,4-cyclohexadiene //
In the reaction below, what are the IUPAC names of the reactant and the product, respectively? (A) methoxybenzene; 1-methoxy-1,4-cyclohexadiene (B) 1-methoxy-1,3,5-cyclohexatriene; 1-methoxy-1,4-cyclohexadiene (C) 1-methylphenol; methyl-1,4-cyclohexadien-4-ol (D) methoxybenzene; 4-methoxy-1,4-cyclohexadiene
