1. The mechanism shown is ___. In Step 2, the carbocation has ___ H’s bonded it.
ID: 902424 • Letter: 1
Question
1. The mechanism shown is ___. In Step 2, the carbocation has ___ H’s bonded it. In Step 3, you would draw a curved arrow from the ___.
SN1
SN2
0
1
2
carbocation to the Cl-
Cl- to H
Cl- to carbocation
2. 1-butanol --> 1-butene is a _____ reaction.
substitution
elimination
proton transfer
addition
3. 2-chloro-2-pentene has ____ stereoisomers and 4-chloro-1-pentene has ____ stereoisomers.
E and Z/cis and trans
no/no
E and Z/no
4. 2-chloro-2-pentene is ____ stable than 4-chloro-2-pentene.
more
less
same
5. What is not needed by a reactant to undergo an elimination reaction?
leaving group
? carbon
H bonded to a ? carbon
pi bond
nucleophile
For Question 1: Step 1 OHH-CI OH2 + CI OH2 OH2 +Cl+H2O +CI H2 to Step 3 CIH2O C H20Explanation / Answer
The mechanism shown is SN1 because SN1 proceeds stepwise in which the leaving group first leaves after that a carbocation forms which is later attacked by the nucleophile. In step 2 the Carbocation formed has 0 H atom bonded because it is 3 degree Carbocation. In step 3 we draw a curved arrow from Cl- to Carbocation because it attacks the Carbocation.
2.
1-butanol to butene is Elimination reaction due to the elimination of OH group to form a double bond
3.
2-chloro-2-pentene has E and Z stereoisomers but 4-chloro-1-pentene has no stereoisomers
4.
2-chloro-2-pentene is less stable than 4-chloro-1-pentene because 4-chloro-1-pentene has more stereoisomers which leads to higher stability.
5.
Pi bond is not needed by a reactant to undergo an elimination reaction because the elimination reaction will lead to the formation of pi bond
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