Alkyl and Aryl Halide MCQ Questions & Answers in Organic Chemistry | Chemistry

A compound having two chiral centres can exist in 4 stereoisomeric forms $$\left( {{2^n}} \right).$$

$${\left( {C{H_3}} \right)_3}C - MgCl + {D_2}O \to {\left( {C{H_3}} \right)_3}C - D + Mg\left( {OD} \right)Cl$$

It is an alkene, ( i.e. contains double bond between carbon atoms ) so the bond angle is $${120^ \circ }$$  while tetrachloro methane is an alkane, i.e. contains only single bonds, so the bond angle is $${109^ \circ }28'.$$

Chlorination beyond monochlorination during the preparation of alkyl halides in presence of $$UV$$  light can be suppressed by taking alkane in excess.
NOTE : For isomeric alkanes the one having largest straight chain has highest b.p. because of large surface area.

\[\underset{\begin{smallmatrix} \text{Methyl} \\ \text{bromide} \end{smallmatrix}}{\mathop{C{{H}_{3}}Br}}\,\xrightarrow[-KBr]{KCN}\underset{\begin{smallmatrix} \text{Ethane} \\ \text{nitrile} \\ A \end{smallmatrix}}{\mathop{C{{H}_{3}}CN}}\,\xrightarrow[\begin{smallmatrix} \text{Complate} \\ \text{hydrolysis} \end{smallmatrix}]{{{H}_{3}}{{O}^{+}}}\underset{\begin{smallmatrix} \text{Ethanoic} \\ \text{acid} \\ B \end{smallmatrix}}{\mathop{C{{H}_{3}}COOH}}\,\xrightarrow[\text{Ether}]{LiAl{{H}_{4}}}\underset{\begin{smallmatrix} \text{Ethanol} \\ \,\,\,\,\,\,\,\,C \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,\]
In the presence of $$LiAl{H_4}$$  carboxylic acid reduce in alcohols directly.

In aromatic system all the carbon and hydrogen atoms are present in one plane and all $$C-C$$  bond lengths are same due to resonance and all carbon atoms have $$s{p^2}$$ hybridisation, so the bond angle is $${120^ \circ }.$$

\[C{{H}_{3}}C{{H}_{2}}Cl\xrightarrow{NaCN}C{{H}_{3}}-\underset{X}{\mathop{C{{H}_{2}}}}\,-CN\]         \[\xrightarrow{\frac{Ni}{{{H}_{2}}}}C{{H}_{3}}-\underset{Y}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}N{{H}_{2}}\xrightarrow[\text{anhydride}]{\text{Acetic}}\]         \[C{{H}_{3}}-C{{H}_{2}}-\underset{Z}{\mathop{C{{H}_{2}}}}\,-NHCOC{{H}_{3}}\]

Chloroform is slowly oxidised into a poisonous compound called phosgene in the presence of air or light. This compound is also called carbonyl chloride $$\left( {COC{l_2}} \right)$$
\[CHC{{l}_{3}}+\frac{1}{2}{{O}_{2}}\xrightarrow[\text{light}]{\text{Air}}\underset{\begin{smallmatrix} \text{Carbonyl chloride} \\ \,\,\,\,\,\,\,\,\text{(phosgene)} \end{smallmatrix}}{\mathop{COC{{l}_{2}}}}\,+HCl\]

Key Idea In chlorobenzene, bromobenzene and chloroethene, lone pair of halogen is delocalised with $$\pi $$ - bonds so it attains double bond character. Thus, these are not suitable as a halide component for Friedel-Crafts reaction.

Other halides, i.e. chloro and bromobenzene along with chloroethene have carbon halogen bond as

The stability of carbocation follow the order $${3^ \circ } > {2^ \circ } > {1^ \circ } > {\text{methyl}}{\text{.}}$$     More the number of alkyl group attached with the carbon atom carrying the positive charge greater would be the tendency to stabilise positive charge via inductive effect and hence more stable is that carbocation.

This carbocation is more stable due to nine $$\alpha $$  - hydrogen and ( nine hyperconjugative structures ) three $$ + I$$  groups.

$${2^ \circ }$$ carbocation containing 6 $$\alpha $$  - hydrogen showing six hyperconjugative structure along with two $$ + I$$  group.

It has slightly lesser stability as compared to $${3^ \circ }$$ - alkyl carbocation due to presence of three electron donating alkyl group in $${3^ \circ }$$ - alkyl carbocation. Although the stabilities of $${3^ \circ }$$ and benzyl carbonium ion are almost same and cannot be compared in solution but whenever a comparison is made between Resonance ( the cause of stability in benzyl carbonium ion ) and No bond resonance ( the cause of stability in $${3^ \circ }$$ carbonium ion ) then the former is always preferred hence here in this question benzyl carbonium ion is more stable than $${3^ \circ }$$  carbonium ion.

$${1^ \circ }$$ carbocation less stable than all present here