Give the products of the following reactions a) CH3-CH2-CH2-OCH3 +HBr b) C6H5OC2H5 +HBr c) (CH3)3COC2H5 +HI d) C6H5OC2H 5 conc HNO3+conc H2SO4 e) CH3CH=CH2 H3O+ f) CH3-CH2-CH-CHO | NaBH4 CH3
Dear student,
The products of the reactions are as follows:
a) CH3-CH2-CH2-OCH3 +HBr CH3-CH2-CH2-Br + CH3OH
In this reaction, C-O bond is broken and is replaced by C-Br bond.
b) C6H5OC2H5 +HBr C6H5OH + C2H5Br
In this reaction, the O-C2H5 bond is broken and phenol is formed as the final product.
c) (CH3)3COC2H5 +HI (CH3)3C-I + C2H5OH
Tertiary ether when reacted with halogen hydride results in the formation of tertiary carbocation , which is highly stable. This carbocation finally reacts with HI to form tertiary iodide.
d) C6H5OC2H5 + conc HNO3 + conc H2SO4
This is the nitration of ethoxybenzene in which two products are formed. The two products are o- and p- ethoxybenzene.
The major product of the two is the para isomer because when -NO2 is present at ortho position, then steric hindrance causes problem and reduces the stability of the compound formed.
e) CH3CH=CH2 + H3O+ CH3CH(OH)CH3 + H2O
Propene when reacted with hydronium ions results in the formation of secondary alcohol. The side product formed is water.
f) CH3-CH2-CH-CHO + NaBH4 CH3-CH2-CH2-CH2OH
Butanal when reacts with NaBH4 forms butanol. NaBH4 is a reducing agent which converts carbonyl to alcohol.
Regards
The products of the reactions are as follows:
a) CH3-CH2-CH2-OCH3 +HBr CH3-CH2-CH2-Br + CH3OH
In this reaction, C-O bond is broken and is replaced by C-Br bond.
b) C6H5OC2H5 +HBr C6H5OH + C2H5Br
In this reaction, the O-C2H5 bond is broken and phenol is formed as the final product.
c) (CH3)3COC2H5 +HI (CH3)3C-I + C2H5OH
Tertiary ether when reacted with halogen hydride results in the formation of tertiary carbocation , which is highly stable. This carbocation finally reacts with HI to form tertiary iodide.
d) C6H5OC2H5 + conc HNO3 + conc H2SO4
This is the nitration of ethoxybenzene in which two products are formed. The two products are o- and p- ethoxybenzene.
The major product of the two is the para isomer because when -NO2 is present at ortho position, then steric hindrance causes problem and reduces the stability of the compound formed.
e) CH3CH=CH2 + H3O+ CH3CH(OH)CH3 + H2O
Propene when reacted with hydronium ions results in the formation of secondary alcohol. The side product formed is water.
f) CH3-CH2-CH-CHO + NaBH4 CH3-CH2-CH2-CH2OH
Butanal when reacts with NaBH4 forms butanol. NaBH4 is a reducing agent which converts carbonyl to alcohol.
Regards