Question

All the compounds which gives positive test with Bredy's reagent (2,4-DNP) and are having molecular mass 100 are treated separately with $NaBH_4$. Product of how many of these can be used effectively to separate racemic mixture of 2-methylbutanoic acid.

Answer 1

8

Answer 2

The problem asks for the number of carbonyl compounds with molecular mass 100 that give a positive test with Bredy's reagent (2,4-DNP) and whose reduction product with $NaBH_4$ can be used effectively to separate a racemic mixture of 2-methylbutanoic acid.

First, identify the carbonyl compounds with molecular formula $C_6H_{12}O$ (molecular mass $6 \times 12 + 12 \times 1 + 1 \times 16 = 72 + 12 + 16 = 100$). These are aldehydes and ketones. All aldehydes and ketones give a positive test with Bredy's reagent.

The possible aldehydes ($R-CHO$) with formula $C_6H_{12}O$ have $R = C_5H_{11}$.

1. Hexanal ($CH_3CH_2CH_2CH_2CH_2CHO$)
2. 4-Methylpentanal ($(CH_3)_2CHCH_2CH_2CHO$)
3. 3-Methylpentanal ($CH_3CH_2CH(CH_3)CHO$) - Chiral
4. 2,2-Dimethylbutanal ($CH_3CH_2C(CH_3)_2CHO$)
5. 3,3-Dimethylbutanal ($(CH_3)_3CCH_2CHO$)
6. 2-Ethylbutanal ($(CH_3CH_2)_2CHCHO$) - Chiral
7. 2,3-Dimethylbutanal ($(CH_3)_2CHCH(CH_3)CHO$) - Chiral

The possible ketones ($R-CO-R'$) with formula $C_6H_{12}O$ have $R+R' = C_5H_{11}$.

1. 2-Hexanone ($CH_3COCH_2CH_2CH_2CH_3$)
2. 4-Methyl-2-pentanone ($CH_3COCH_2CH(CH_3)_2$)
3. 3-Methyl-2-pentanone ($CH_3COCH(CH_3)CH_2CH_3$) - Chiral
4. 3,3-Dimethyl-2-butanone ($CH_3COC(CH_3)_3$)
5. 3-Hexanone ($CH_3CH_2COCH_2CH_2CH_3$)
6. 2-Methyl-3-pentanone ($CH_3CH_2COCH(CH_3)_2$)

Total carbonyl compounds = 7 aldehydes + 6 ketones = 13.

Next, these compounds are reduced with $NaBH_4$. Aldehydes are reduced to primary alcohols, and ketones are reduced to secondary alcohols. $NaBH_4$ is an achiral reducing agent.

To separate a racemic mixture of 2-methylbutanoic acid (a chiral acid), one common method is to react it with a pure chiral alcohol to form diastereomeric esters, which can then be separated. This requires the alcohol used for separation to be a pure enantiomer of a chiral alcohol.

When a carbonyl compound is reduced by $NaBH_4$:

• If the starting carbonyl compound is achiral, the resulting alcohol will be achiral. Achiral alcohols cannot be used as chiral resolving agents.
• If the starting carbonyl compound is chiral, it will exist as a racemic mixture (unless specified otherwise). Reduction of a racemic chiral carbonyl compound with an achiral reagent like $NaBH_4$ will produce a racemic mixture of the corresponding chiral alcohol. A racemic alcohol cannot be used as a pure chiral resolving agent.
• If the starting carbonyl compound is prochiral (achiral but can be converted into a chiral molecule upon reaction at the prochiral center), reduction with an achiral reagent like $NaBH_4$ will produce a racemic mixture of the resulting chiral alcohol.

Therefore, for the reduction product (alcohol) to be a pure chiral alcohol suitable for separating a racemic acid, the starting carbonyl compound must be prochiral, and the reduction must be carried out enantioselectively using a chiral reducing agent (like CBS reagent or $LiAlH_4$ with chiral ligands), or the resulting racemic alcohol must be resolved into its pure enantiomers by some other method.

However, the question specifies reduction with $NaBH_4$, which is an achiral reagent. This means the reduction of any prochiral or chiral carbonyl compound will yield a racemic mixture of the corresponding alcohol. The reduction of an achiral carbonyl compound will yield an achiral alcohol.

Let's examine the reduction products and their chirality when reduced with $NaBH_4$:

Aldehydes ($R-CHO \xrightarrow{NaBH_4} R-CH_2OH$):

1. Hexanal $\rightarrow$ 1-Hexanol (Achiral)
2. 4-Methylpentanal $\rightarrow$ 4-Methyl-1-pentanol (Achiral)
3. 3-Methylpentanal (Chiral) $\rightarrow$ (R/S)-3-Methyl-1-pentanol (Racemic chiral alcohol)
4. 2,2-Dimethylbutanal $\rightarrow$ 2,2-Dimethyl-1-butanol (Achiral)
5. 3,3-Dimethylbutanal $\rightarrow$ 3,3-Dimethyl-1-butanol (Achiral)
6. 2-Ethylbutanal (Chiral) $\rightarrow$ (R/S)-2-Ethyl-1-butanol (Racemic chiral alcohol)
7. 2,3-Dimethylbutanal (Chiral, 2 chiral centers) $\rightarrow$ Mixture of diastereomers (R,R/S,S and R,S/S,R) of 2,3-Dimethyl-1-butanol. The product is a racemic mixture of chiral compounds.

Ketones ($R-CO-R' \xrightarrow{NaBH_4} R-CH(OH)-R'$):

1. 2-Hexanone (Prochiral) $\rightarrow$ (R/S)-2-Hexanol (Racemic chiral alcohol)
2. 4-Methyl-2-pentanone (Prochiral) $\rightarrow$ (R/S)-4-Methyl-2-pentanol (Racemic chiral alcohol)
3. 3-Methyl-2-pentanone (Chiral) $\rightarrow$ Mixture of diastereomers (e.g., R-ketone gives R,R and S,R alcohols; S-ketone gives R,S and S,S alcohols) of 3-Methyl-2-pentanol. The product is a mixture of stereoisomers, including enantiomers, resulting in a racemic mixture of the chiral alcohol(s).
4. 3,3-Dimethyl-2-butanone (Achiral) $\rightarrow$ 3,3-Dimethyl-2-butanol (Achiral)
5. 3-Hexanone (Prochiral) $\rightarrow$ (R/S)-3-Hexanol (Racemic chiral alcohol)
6. 2-Methyl-3-pentanone (Prochiral) $\rightarrow$ (R/S)-2-Methyl-3-pentanol (Racemic chiral alcohol)

The reduction products are either achiral alcohols or racemic mixtures of chiral alcohols (or mixtures of stereoisomers resulting in a racemic mixture overall). None of these are pure chiral alcohols.

The question asks how many of these (referring to the initial 13 carbonyl compounds) have a reduction product that can be used effectively to separate a racemic mixture. This implies that the reduction product itself, as obtained from the $NaBH_4$ reduction, is the resolving agent. Since $NaBH_4$ produces either achiral alcohols or racemic chiral alcohols, and a pure chiral alcohol is required for effective resolution via diastereomer formation, it seems that none of the direct reduction products can be used effectively as a pure chiral resolving agent in the standard manner.

However, the question might be interpreted differently. It might imply that the structure of the alcohol produced is suitable for resolution, and that the alcohol can be obtained in a pure enantiomeric form (perhaps via resolution of the racemic alcohol itself, or by starting with a pure enantiomer of the carbonyl compound if it's chiral, or by using asymmetric reduction). Given the context of typical JEE/NEET questions and the wording "product of how many of these", it is more likely that the question expects us to identify which starting materials yield alcohols that are chiral, assuming that these chiral alcohols could be obtained in pure form (even if the $NaBH_4$ reduction itself doesn't produce them in pure form). If the alcohol product is achiral, it definitely cannot be used.

Let's re-evaluate based on this interpretation: Which of the 13 carbonyl compounds produce a chiral alcohol upon reduction?

• Aldehydes: 3-Methylpentanal, 2-Ethylbutanal, 2,3-Dimethylbutanal produce chiral primary alcohols. (3 compounds)
• Ketones: 2-Hexanone, 4-Methyl-2-pentanone, 3-Methyl-2-pentanone, 3-Hexanone, 2-Methyl-3-pentanone produce chiral secondary alcohols. (5 compounds)

Total number of carbonyl compounds whose reduction product is a chiral alcohol = 3 + 5 = 8.

These 8 alcohols are:

1. 3-Methyl-1-pentanol ($CH_3CH_2CH(CH_3)CH_2OH$) - Chiral at C3
2. 2-Ethyl-1-butanol ($(CH_3CH_2)_2CHCH_2OH$) - Chiral at C2
3. 2,3-Dimethyl-1-butanol ($(CH_3)_2CHCH(CH_3)CH_2OH$) - Chiral at C2 and C3
4. 2-Hexanol ($CH_3CH(OH)CH_2CH_2CH_2CH_3$) - Chiral at C2
5. 4-Methyl-2-pentanol ($CH_3CH(OH)CH_2CH(CH_3)_2$) - Chiral at C2
6. 3-Methyl-2-pentanol ($CH_3CH(OH)CH(CH_3)CH_2CH_3$) - Chiral at C2 and C3
7. 3-Hexanol ($CH_3CH_2CH(OH)CH_2CH_2CH_3$) - Chiral at C3
8. 2-Methyl-3-pentanol ($CH_3CH_2CH(OH)CH(CH_3)_2$) - Chiral at C3

All these 8 compounds yield chiral alcohols upon reduction. If these chiral alcohols were available in pure enantiomeric form, they could be used to resolve racemic 2-methylbutanoic acid by forming diastereomeric esters.

Given the question phrasing, it is most likely asking for the number of starting carbonyl compounds that can yield a chiral alcohol upon reduction, implying that this chiral alcohol can then be used for separation (assuming it can be obtained in pure form).

The number of such compounds is 8.

Let's double-check the chiral centers in the alcohols:

1. 3-Methyl-1-pentanol: $CH_3CH_2\stackrel{*}{C}H(CH_3)CH_2OH$ (C3 is chiral)
2. 2-Ethyl-1-butanol: $CH_3CH_2\stackrel{*}{C}H(CH_2CH_3)CH_2OH$ (C2 is chiral)
3. 2,3-Dimethyl-1-butanol: $(CH_3)_2\stackrel{*}{C}H\stackrel{*}{C}H(CH_3)CH_2OH$ (C2 and C3 are chiral)
4. 2-Hexanol: $CH_3\stackrel{*}{C}H(OH)CH_2CH_2CH_2CH_3$ (C2 is chiral)
5. 4-Methyl-2-pentanol: $CH_3\stackrel{*}{C}H(OH)CH_2CH(CH_3)_2$ (C2 is chiral)
6. 3-Methyl-2-pentanol: $CH_3\stackrel{*}{C}H(OH)\stackrel{*}{C}H(CH_3)CH_2CH_3$ (C2 and C3 are chiral)
7. 3-Hexanol: $CH_3CH_2\stackrel{*}{C}H(OH)CH_2CH_2CH_3$ (C3 is chiral)
8. 2-Methyl-3-pentanol: $CH_3CH_2\stackrel{*}{C}H(OH)\stackrel{*}{C}H(CH_3)_2$ (C3 is chiral)

All 8 identified alcohols are indeed chiral. Therefore, the 8 corresponding carbonyl compounds are the ones whose reduction products (chiral alcohols) could be used for resolution if obtained in pure form.

Final list of the 8 carbonyl compounds:

Aldehydes:

1. 3-Methylpentanal
2. 2-Ethylbutanal
3. 2,3-Dimethylbutanal

Ketones:

1. 2-Hexanone
2. 4-Methyl-2-pentanone
3. 3-Methyl-2-pentanone
4. 3-Hexanone
5. 2-Methyl-3-pentanone

The number of such compounds is 8.

The final answer is $\boxed{8}$.