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Question: Which of the following is not correct about the reaction, \( C{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\xright...

Which of the following is not correct about the reaction, CH2=CH2+Br2→(aq)NaIC{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\xrightarrow[\left( aq \right)]{NaI}
(A) The products formed CH2BrCH2BrC{{H}_{2}}BrC{{H}_{2}}Br are and CH2BrCH2I.C{{H}_{2}}BrC{{H}_{2}}I.
(B) The reaction follows a polar mechanism.
(C) The reaction occurs readily in solution and is catalyzed by inorganic halides.
(D) CH2ICH2IC{{H}_{2}}IC{{H}_{2}}I is formed only.

Explanation

Solution

Hint : As we know that the alkenes are very reactive towards electrophilic substitution reaction. The reactivity is due to its electrons present in it. All the halides are electronegative in nature and can act as a nucleophile.

Complete Step By Step Answer:
We should be well aware of the fact that in such a type of reaction the amount of the attacking component decides the fate of the reaction and the product formation. Also Bromine is considered as enough of a reducing agent, when amount is sufficient it will reduce the reactant into product and results in colour formation.
The ethene molecule contains a double bond, one is sigma bonding and the other is pi bonding. The bonding electrons are not very stable due to side-wise overlapping. Therefore, the pi electrons are responsible for the reactivity of ethene compounds toward electrophilic substitution. When the ethene attacks on bromine, the atom’s shared pair electrons repel with ethene pi electrons and then one bromine atom splits into bromide ion and one is bromonium ion. This bromonium ion is very electron deficient; the pi electrons on each carbon of ethene are polarized by positive charge on bromine and hence cyclization occurs with the ethene compound.
CH2=CH2+Br2→(aq)NaII−CH2−CH2−IProduct I  C{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\xrightarrow[\left( aq \right)]{NaI}\underset{Product\text{ }I}{\mathop{I-CH2-CH2-I}}\,\text{ }
But the reaction will also undergo polar mechanism;
CH2=CH2+Br2→(aq)NaICH2+−CH2−BrC{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\xrightarrow[\left( aq \right)]{NaI}CH_{2}^{+}-C{{H}_{2}}-Br
Now the carbocation will be attacked by a nucleophile;
CH2+−CH2−Br→(aq)NaII−CH2−CH2−Br Product II CH_{2}^{+}-C{{H}_{2}}-Br\xrightarrow[\left( aq \right)]{NaI}\underset{Product\text{ }II}{\mathop{I-C{{H}_{2}}-C{{H}_{2}}-Br\text{ }}}\,
Therefore, the correct answer is option A.

Note :
Remember that there are two types of state of electrons; Ground state- in which electrons are paired up. The excited state -in which electrons are in triplet state. The ethene electrons when reacted with bromonium ions are excited to triplet state then the cyclization occurs.