
$PhCH = C{H_2}\xrightarrow[{C{H_2}C{l_2}}]{{ArC{O_3}H}}A\xrightarrow[{2.{H_2}O}]{{1.LiAl{H_4}.E{t_2}O}}B$
The end product B is?
Answer
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Hint: Most epoxides are generated by treating alkenes with peroxide-containing reagents, which donate a single oxygen atom. Safety considerations weigh on these reactions because organic peroxides are prone to spontaneous decomposition or even combustion. Lithium aluminium hydride ($LiAl{H_4}$ ) is a strong reducing agent and can reduce the epoxide to secondary alcohol.
Complete answer:
The conversion of the reactant into A is a specific reaction in organic chemistry, which is known as the Prilezhaev reaction. The reaction proceeds via what is commonly known as the "Butterfly Mechanism". The peroxide is viewed as an electrophile, and the alkene a nucleophile. The reaction is considered to be concerted (the numbers in the mechanism below are for simplification). The butterfly mechanism allows ideal positioning of the\[O - O\] sigma star orbital for \[C - C\] pi-electrons to attack because two bonds are broken and formed to the epoxide oxygen; this is formally an example of a coarctate transition state. The equation for the reaction is as follows:
$PhCH = C{H_2}\xrightarrow[{C{H_2}C{l_2}}]{{ArC{O_3}H}}{C_8}{H_8}O$
The product (A) formed above is known as 2-phenyloxirane.
In the next reaction, we find a strong reducing agent known as lithium aluminium hydride ($LiAl{H_4}$ ). When 2-phenyloxirane, which is the product (A) is treated with $LiAl{H_4}$in the aqueous medium, a secondary alcohol is formed. The water is used to absorb the heat released from the breakage of the epoxide bond. The reaction is as follows:
${C_8}{H_8}O\xrightarrow[{2.{H_2}O}]{{1.LiAl{H_4}}}PhCH(OH)C{H_3}$
The product (B) formed above is (1-hydroxy ethyl) benzene.
Thus, the overall reaction is as follows:
$PhCH = C{H_2}\xrightarrow[{C{H_2}C{l_2}}]{{ArC{O_3}H}}{C_8}{H_8}O(A)\xrightarrow[{2.{H_2}O}]{{1.LiAl{H_4}}}PhCH(OH)C{H_3}(B)$
Note:
A compound containing the epoxide functional group can be called an epoxy, epoxide, oxirane, and ethoxyline. Simple epoxides are often referred to as oxides. Thus, the epoxide of ethylene (\[{C_2}{H_4}\] ) is ethylene oxide (\[{C_2}{H_4}O\] ). Many compounds have trivial names; for instance, ethylene oxide is called "oxirane".
Complete answer:
The conversion of the reactant into A is a specific reaction in organic chemistry, which is known as the Prilezhaev reaction. The reaction proceeds via what is commonly known as the "Butterfly Mechanism". The peroxide is viewed as an electrophile, and the alkene a nucleophile. The reaction is considered to be concerted (the numbers in the mechanism below are for simplification). The butterfly mechanism allows ideal positioning of the\[O - O\] sigma star orbital for \[C - C\] pi-electrons to attack because two bonds are broken and formed to the epoxide oxygen; this is formally an example of a coarctate transition state. The equation for the reaction is as follows:
$PhCH = C{H_2}\xrightarrow[{C{H_2}C{l_2}}]{{ArC{O_3}H}}{C_8}{H_8}O$
The product (A) formed above is known as 2-phenyloxirane.
In the next reaction, we find a strong reducing agent known as lithium aluminium hydride ($LiAl{H_4}$ ). When 2-phenyloxirane, which is the product (A) is treated with $LiAl{H_4}$in the aqueous medium, a secondary alcohol is formed. The water is used to absorb the heat released from the breakage of the epoxide bond. The reaction is as follows:
${C_8}{H_8}O\xrightarrow[{2.{H_2}O}]{{1.LiAl{H_4}}}PhCH(OH)C{H_3}$
The product (B) formed above is (1-hydroxy ethyl) benzene.
Thus, the overall reaction is as follows:
$PhCH = C{H_2}\xrightarrow[{C{H_2}C{l_2}}]{{ArC{O_3}H}}{C_8}{H_8}O(A)\xrightarrow[{2.{H_2}O}]{{1.LiAl{H_4}}}PhCH(OH)C{H_3}(B)$
Note:
A compound containing the epoxide functional group can be called an epoxy, epoxide, oxirane, and ethoxyline. Simple epoxides are often referred to as oxides. Thus, the epoxide of ethylene (\[{C_2}{H_4}\] ) is ethylene oxide (\[{C_2}{H_4}O\] ). Many compounds have trivial names; for instance, ethylene oxide is called "oxirane".
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