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26 Nov 2019

Please explain

1. In electrophilic aromatic substitution, why does an amide groupdirect substitution to a position opposite to it on the aromaticring?
a.) The N donates electron density to the ring by induction anddestabilizes the intermediate that forms meta to it
b.) The N donates electron density to the ring by induction andstabilizes the intermediate that forms para to it
c.) The N donates electron density to the ring by resonance anddestabilizes the intermediate that forms meta to it
d.) The N donates electron density to the ring by resonance andstabilizes the intermediate that forms para to it
e.) The N withdraws electron density from the ring by induction anddestabilizes the intermediate that forms meta to it

2. Which of the following cannot be made by the reduction of analdehyde or ketone with NaBH4?
a.) 1-butanol
b.) 2-butanol
c.) 1-methyl-2-butanol
d.) 2-methyl-2-butanol
e.) all these alcohols can be made by NaBH4 reduction

3. Which of the following represents the correct order in terms ofboiling point?
a.) n-pentane < 1-butanol < diethyl ether <2-butanone
b.) n-pentane < 2-butanone < diethyl ether <1-butanol
c.) 2-butanone < n-pentane < diethyl ether <1-butanol
d.) n-pentane < diethyl ether < 1-butanol <2-butanone
e.) none of the above

4. Which of the following undergoes SN2 substitution with CH3O-1most rapidly?
a.) PhCH2Br
b.) Ph3CBr
c.) PhCH2CH2Br
d.) PhBr
e.) PhCH2CH2CH2Br

5.Which of the following is true about the nitration ofanisole?
a.) It proceeds more rapidly than the nitration of benzene andpredominantly yields the meta product
b.) It proceeds slower than the nitration of benzene andpredominantly yields the meta product
c.) It proceeds slower than the nitration of benzene andpredominantly yields the ortho and para product
d.) It proceeds more rapidly than the nitration of benzene andpredominantly yields the otho and para products
e.) It proceeds at the same rate as the nitration of benzene andpredominantly yields the meta product

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Deanna Hettinger
Deanna HettingerLv2
18 Mar 2019

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