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Question

(B) The reaction involves the following steps:$1$. The reaction of $3$-methylbutan-$2$-one (an aldehyde/ketone) with methylamine $(CH_3NH_2)$ forms an

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$(CH_3)_2CH-CH_2-NHCH_3$

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(B) The reaction involves the following steps:$1$. The reaction of $3$-methylbutan-$2$-one (an aldehyde/ketone) with methylamine $(CH_3NH_2)$ forms an imine intermediate $(R_2C=NCH_3)$.$2$. The subsequent reduction with $LiAlH_4$ reduces the $C=N$ double bond to a $C-NH$ single bond,resulting in a secondary amine.$3$. The starting material is $3$-methylbutan-$2$-one,which is $(CH_3)_2CH-C(=O)-CH_3$.$4$. The reaction with $CH_3NH_2$ forms the imine $(CH_3)_2CH-C(=NCH_3)-CH_3$.$5$. Reduction with $LiAlH_4$ gives the amine $(CH_3)_2CH-CH(NHCH_3)-CH_3$.

The major organic product formed from the following reaction is:

Similar Questions

Identify what is $Y$ in the following reaction sequence?
$CH_3CONH_2$ $\xrightarrow{Br_2/NaOH(aq)} X$ $\xrightarrow[(ii) H_2O]{(i) NaNO_2 + HCl} Y$

The incorrect statement regarding the reaction given below is:

Suppose the following reaction:
$CH_3-CH(CH_3)-COOH + CH_3-NH_2$ $\rightarrow 'A'$ $\xrightarrow{\Delta} 'B'$ $\xrightarrow{LiAlH_4 (excess)} 'C'$
The final product '$C$' will be:

Reaction of aniline with $NaNO_{2} +$ dil. $HCl$ at $0^{\circ} C$ followed by reaction with $CuCN$ yields:

$Br-CH_2-(CH_2)_2-CH_2-Br + CH_3NH_2 \to$ The product of the reaction is:

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