An alkane (molecular weight 72) forms only on | vedclass

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Question

(A) The general formula for an alkane is $C_nH_{2n+2}$.Given molecular weight is $72$.$12n + 1(2n+2) = 72 \implies 14n + 2 = 72 \implies 14n = 70 \imp

Vedclass · Accepted Answer

$(CH_3)_4C$

Vedclass · Answer

(A) The general formula for an alkane is $C_nH_{2n+2}$.Given molecular weight is $72$.$12n + 1(2n+2) = 72 \implies 14n + 2 = 72 \implies 14n = 70 \implies n = 5$.So,the molecular formula is $C_5H_{12}$.An alkane that forms only one monochlorinated product must have all its hydrogen atoms equivalent.In $2,2-$dimethylpropane (neopentane),$(CH_3)_4C$,all $12$ hydrogen atoms are equivalent.Therefore,it forms only one monochlorinated product upon reaction with $Cl_2$ in the presence of sunlight.

An alkane (molecular weight $72$) forms only one monochlorinated product. Its formula is

Similar Questions

Different hydrogen atoms in the compound are represented by alphabets:
$CH_3(A)-CH=CH(B)-CH_2(C)-CH_2(D)-CH(E)(CH_3)_2(F)$
Arrange them in decreasing order of reactivity towards free radical substitution.

Considering the following reaction,what is the structure of the major product '$X$'?
$H_3C-CH(D)-CH(CH_3)-CH_3 + Br^\bullet \to 'X' + HBr$

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$(a)$ Hydrolysis of $Al_4C_3$
$(b)$ Hydrolysis of $Be_2C$
$(c)$ Hydrolysis of $Mg_2C_3$
$(d)$ Hydrolysis of $CaC_2$

What is the major product?
$(CH_3-CH_2)_2CuLi \xrightarrow{CH_3-Cl} \text{Product}$

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