Looking for classes? Ksquare Career Institute, Bengaluru →
The major organic product formed from cyclohexanone in the following reaction
Cyclohexanone $\xrightarrow{(i)\ CH_3NH_2}$ $\xrightarrow{(ii)\ LiAlH_4}$ $\xrightarrow{(iii)\ H_2O}$ product
is -
[add image]
Cyclohexanone $\xrightarrow{(i)\ CH_3NH_2}$ $\xrightarrow{(ii)\ LiAlH_4}$ $\xrightarrow{(iii)\ H_2O}$ product
is -
[add image]
A
Structure 1: a ring carbon carrying both $-OH$ and $-NHCH_3$ (drawn with H on N) [add image]
B
N-methylcyclohexylamine, $C_6H_{11}-NHCH_3$
C
Structure 3: a ring carbon carrying both $-OH$ and $-NHCH_3$ (drawn with H above N) [add image]
D
$C_6H_{11}-O-NHCH_3$
Detailed Solution
Step 1: methylamine, a primary amine, adds to the carbonyl group of cyclohexanone. The addition product (a carbinolamine) loses a molecule of water to form an imine (a Schiff base).
$C_6H_{10}=O + CH_3NH_2 \rightarrow C_6H_{10}=N-CH_3 + H_2O$ (N-methylcyclohexanimine)
Step 2: $LiAlH_4$ reduces the C=N double bond of the imine.
Step 3: hydrolysis (work-up with water) gives the secondary amine.
$C_6H_{10}=N-CH_3 \xrightarrow{LiAlH_4,\ H_2O} C_6H_{11}-NH-CH_3$
The carbinolamine is only an unstable intermediate and is not the final product, and no N-O bond is formed.
So the major product is N-methylcyclohexylamine (N-methylcyclohexanamine).
$C_6H_{10}=O + CH_3NH_2 \rightarrow C_6H_{10}=N-CH_3 + H_2O$ (N-methylcyclohexanimine)
Step 2: $LiAlH_4$ reduces the C=N double bond of the imine.
Step 3: hydrolysis (work-up with water) gives the secondary amine.
$C_6H_{10}=N-CH_3 \xrightarrow{LiAlH_4,\ H_2O} C_6H_{11}-NH-CH_3$
The carbinolamine is only an unstable intermediate and is not the final product, and no N-O bond is formed.
So the major product is N-methylcyclohexylamine (N-methylcyclohexanamine).