Organic Question about sn12/e12

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Slaughter421

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n-propyl chloride ---->KOH(alc) to make what?

The answer is propylene.

My book says this is an E1 reaction, but my question is why cant this be a sn1.

I know SN1 wants 3>2>1 but doesnt e1 want the same?

Why not sn2 occur or e2 occur? please explain why.
What is the alc mean?

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n-propyl chloride ---->KOH(alc) to make what?

The answer is propylene.

My book says this is an E1 reaction, but my question is why cant this be a sn1.

I know SN1 wants 3>2>1 but doesnt e1 want the same?

Why not sn2 occur or e2 occur? please explain why.
What is the alc mean?

the KOH is a very strong BASE, bases favor elimination rxns. if it was KCN
it might compete with sn2 if the chlorine was on a primary (terminal) carbon. also Cl is a poor leaving group, and the "1" rxns need good leaving groups
 
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n-propyl chloride ---->KOH(alc) to make what?

The answer is propylene.

My book says this is an E1 reaction, but my question is why cant this be a sn1.

I know SN1 wants 3>2>1 but doesnt e1 want the same?

Why not sn2 occur or e2 occur? please explain why.
What is the alc mean?

I don’t really know where the propylene product is came from. 😕
n-propyl chloride is a “primary” alkyl halide and KOH is a strong base (strong bases act as strong nucleophiles when the substrate is primary such as in this case) ; Therefore the reaction is “SN2 backside attack” and the product is (n-propyl alcohol). I really think that the propylene answer is wrong!
Try to do the reaction step by step by yourself and start by drawing a pair of electrons above and below the oxygen in the KOH base, then extend an arrow from the Hydroxide oxygen and attack behind the alpha C , which will blow off the Cl. At the end of the reaction the OH is going to replace the Cl.


Hope that helps
 
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the alcoholic presence of KOH actually forms KO- ions in the solution which is a very strong base. The aqueos solution of KOH is ionized with K+ and OH- which is a strong base but also a very strong nucleophile, in this case substitution would be the preferred mechanism.

In the alcholic KOH KO- is the strong base, and elimination is preferred because the nucleophile is RO-

hope this helps
ps. just memorize it dont apply it
 
the alcoholic presence of KOH actually forms KO- ions in the solution which is a very strong base. The aqueos solution of KOH is ionized with K+ and OH- which is a strong base but also a very strong nucleophile, in this case substitution would be the preferred mechanism.

In the alcholic KOH KO- is the strong base, and elimination is preferred because the nucleophile is RO-

hope this helps
ps. just memorize it dont apply it

Im kinda understanding now. This is basically an exception to the rule.

The rule being that if a reactant is a strong nuc and a strong base (RO-) then methyl and primary will go to SN2 while secondary and tertiary will go to e2.

You said KO- was a strong base as well as a strong nuc. Based off the rule it would go SN2. So is this an exception or did u mean KO- was just a strong base.
 
the alcoholic presence of KOH actually forms KO- ions in the solution which is a very strong base. The aqueos solution of KOH is ionized with K+ and OH- which is a strong base but also a very strong nucleophile, in this case substitution would be the preferred mechanism.

In the alcholic KOH KO- is the strong base, and elimination is preferred because the nucleophile is RO-

hope this helps
ps. just memorize it dont apply it

I guess the nature of substrate and the reagent is what makes the majority contemplate
 
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