Question

Organic compound [M] has the following data:

(i) Molecular weight is 149, it contains only 3 different elements.

(ii) It can rotate plane polarized light (chiral carbon).

(iii) [M] on oxidation with hot alkaline $KMnO_4$ produces benzoic acid.

(iv) [M] can be prepared by treating carbonyl compound with methyl amine followed by $NaBH_4$.

Write the number of possible structures for [M] (excluding stereo isomers).

Answer 1

1

Answer 2

The problem provides information about an organic compound [M]:

(i) Molecular weight = 149, contains only 3 different elements.

(ii) It is chiral.

(iii) Oxidation with hot alkaline $KMnO_4$ produces benzoic acid.

(iv) [M] can be prepared by treating a carbonyl compound with methyl amine followed by $NaBH_4$.

From (iii), the oxidation with $KMnO_4$ yielding benzoic acid indicates that [M] is an alkylbenzene where the alkyl chain attached to the benzene ring is oxidized to a carboxyl group. For this to happen, the carbon directly attached to the benzene ring (benzylic carbon) must have at least one hydrogen, and the entire side chain is cleaved except for the benzylic carbon, which becomes the carboxylic acid carbon. Since the product is benzoic acid ($Ph-COOH$), the benzene ring is unsubstituted except for the side chain. Thus, [M] must be of the form $Ph-R'$, where $R'$ is a group attached to the benzene ring, and the carbon attached to the ring in $R'$ is a $CH$ or $CH_2$ group.

From (iv), [M] is prepared by reductive amination of a carbonyl compound with methyl amine. The general reaction is:

$R_a-C(=O)-R_b + CH_3NH_2 \rightarrow R_a-C(=NCH_3)-R_b \xrightarrow{NaBH_4} R_a-CH(-NHCH_3)-R_b$.

So, [M] is a secondary amine with the structure $R_a-CH(-NHCH_3)-R_b$.

Combining the information from (iii) and (iv):

[M] is a secondary amine $R_a-CH(-NHCH_3)-R_b$ and it contains a phenyl group ($Ph$) attached to the carbon that gets oxidized to -COOH. In the structure $R_a-CH(-NHCH_3)-R_b$, the carbon bearing the amine group is the one derived from the carbonyl carbon. For oxidation to yield benzoic acid, this carbon must be the benzylic carbon, i.e., one of $R_a$ or $R_b$ must be a phenyl group (Ph). Let's assume $R_a = Ph$.

So, the structure of [M] is $Ph-CH(-NHCH_3)-R_b$. The benzylic carbon is the one attached to the phenyl group, which is also the carbon attached to the -NHCH$_3$ group and $R_b$. This carbon has one hydrogen atom (the one shown explicitly in -CH-). Oxidation of this compound will convert the benzylic carbon to -COOH, yielding benzoic acid, provided the rest of the side chain is cleaved. This structure fits the oxidation requirement.

Now let's use the molecular weight information (i). [M] has molecular weight 149 and contains only 3 elements (C, H, N, based on the structure).

The structure is $C_6H_5-CH(-NHCH_3)-R_b$.

The molecular weight of $C_6H_5$ is $6 \times 12 + 5 = 77$.

The fragment $-CH(-NHCH_3)$ has one carbon, one nitrogen, and a methyl group attached to nitrogen. The carbon is attached to the phenyl ring and $R_b$.

The structure is $C_6H_5 - CH(R_b) - NH - CH_3$.

Molecular weight = MW($C_6H_5$) + MW($-CH-$) + MW($R_b$) + MW($-NH-$) + MW($-CH_3$).

No, the structure is $C_6H_5 - CH(R_b) - NHCH_3$.

The atoms are: $C_6H_5$, $C$, $H$, $R_b$, $N$, $CH_3$.

Let's count the atoms based on the formula $C_6H_5-CH(R_b)-NHCH_3$.

Number of carbons = 6 (from $C_6H_5$) + 1 (from -CH-) + number of carbons in $R_b$ + 1 (from -CH$_3$).

Number of hydrogens = 5 (from $C_6H_5$) + 1 (from -CH-) + number of hydrogens in $R_b$ + 1 (from -NH-) + 3 (from -CH$_3$). Note: The explicit hydrogen on the carbon is shown as -CH-. The nitrogen is -NH-.

The formula is $C_6H_5-CH(R_b)-NHCH_3$.

Let $R_b$ be an alkyl group $C_nH_{2n+1}$.

Total carbons = $6 + 1 + n + 1 = 8+n$.

Total hydrogens = $5 + 1 + (2n+1) + 1 + 3 = 11+2n$.

Total nitrogen = 1.

Molecular weight = $12 \times (8+n) + 1 \times (11+2n) + 14 \times 1 = 96 + 12n + 11 + 2n + 14 = 121 + 14n$.

We are given MW = 149.

$121 + 14n = 149$

$14n = 149 - 121 = 28$

$n = 2$.

So, $R_b$ is an alkyl group with 2 carbons, which is ethyl ($C_2H_5$).

The structure of [M] is $Ph-CH(C_2H_5)-NHCH_3$.

Let's check condition (ii): It is chiral.

The structure is $Ph-CH(C_2H_5)-NHCH_3$. The carbon atom attached to the phenyl group, ethyl group, -NHCH$_3$ group, and a hydrogen atom is a chiral center because all four attached groups ($C_6H_5$, $C_2H_5$, $NHCH_3$, H) are different. So, this structure is chiral.

We derived the structure by assuming $R_a = Ph$ and $R_b = C_2H_5$ in the general form $R_a-CH(-NHCH_3)-R_b$. What if $R_b = Ph$ and $R_a = C_2H_5$? The structure would be $C_2H_5-CH(-NHCH_3)-Ph$, which is the same structure $Ph-CH(C_2H_5)-NHCH_3$.

What if the phenyl group is attached elsewhere in the chain? For example, if the carbonyl compound was $Ph-CH_2-C(=O)-R_b$? Reductive amination with methylamine would give $Ph-CH_2-CH(-NHCH_3)-R_b$.

Let's check the molecular weight for $Ph-CH_2-CH(-NHCH_3)-R_b$.

$C_6H_5-CH_2-CH(-NHCH_3)-R_b$.

Let $R_b = C_nH_{2n+1}$.

Carbons = $6 + 1 + 1 + n + 1 = 9+n$.

Hydrogens = $5 + 2 + 1 + (2n+1) + 1 + 3 = 12+2n$.

Nitrogen = 1.

MW = $12(9+n) + (12+2n) + 14 = 108 + 12n + 12 + 2n + 14 = 134 + 14n$.

Setting MW = 149: $134 + 14n = 149 \Rightarrow 14n = 15$. $n = 15/14$, not an integer. So, $R_b$ cannot be an alkyl group in this case.

What if $R_b = H$? Then the carbonyl compound is $Ph-CH_2-CHO$ (phenylacetaldehyde). Reductive amination gives $Ph-CH_2-CH_2-NHCH_3$.

MW = $12 \times 8 + 1 \times 11 + 14 = 96 + 11 + 14 = 121$. Not 149.

Also, oxidation of $Ph-CH_2-CH_2-NHCH_3$ would yield phenylacetic acid ($Ph-CH_2-COOH$), not benzoic acid. So, this class of compounds is not possible.

The only possible structure that fits conditions (i), (iii), and (iv) is $Ph-CH(C_2H_5)-NHCH_3$.

We have already confirmed that this structure is chiral, satisfying condition (ii).

The question asks for the number of possible structures for [M] (excluding stereo isomers). Since we found only one constitutional isomer that satisfies all the conditions, the number of possible structures is 1.

The structure is N-methyl-1-phenylpropan-1-amine.

Carbonyl compound: Propiophenone ($Ph-C(=O)-C_2H_5$)

Amine: Methylamine ($CH_3NH_2$)

Product: $Ph-CH(C_2H_5)-NHCH_3$.

Let's double-check if there are any other possibilities for $R_b$ having MW=29. $R_b$ must be composed of C, H, N. Since it's derived from a carbonyl compound $R_a-C(=O)-R_b$, $R_b$ is typically an alkyl group or H or aryl group. We assumed $R_b$ is an alkyl group $C_nH_{2n+1}$ and found $n=2$. Are there other possibilities for a fragment with MW=29 made of C, H, N?

Possible atoms: C (12), H (1), N (14).

Let the formula be $C_xH_yN_z$. $12x + y + 14z = 29$.

If z=1, $12x + y + 14 = 29 \Rightarrow 12x + y = 15$.

If x=1, $12 + y = 15 \Rightarrow y = 3$. Formula $CH_3$. MW = 15. Not 29.

If x=0, $y = 15$. Formula $H_{15}N$. Impossible.

If z=0, $12x + y = 29$.

If x=1, $12 + y = 29 \Rightarrow y = 17$. Impossible for $C_1$.

If x=2, $24 + y = 29 \Rightarrow y = 5$. Formula $C_2H_5$. MW = 29. This is ethyl.

If x=0, $y = 29$. Impossible.

So, the only fragment with MW=29 made of C and H is $C_2H_5$.

Could $R_b$ contain Nitrogen? If z=1, $12x+y=15$. Possible fragments are $CH_3$ (MW=15).

If z=2, $12x + y + 28 = 29 \Rightarrow 12x + y = 1$. Impossible for non-negative x, y.

So, $R_b$ must be $C_2H_5$.

The only possible constitutional structure for [M] is N-methyl-1-phenylpropan-1-amine.