Organic Question about sn12/e12

[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?

 

[wall1two]

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

 

[wizi]

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?

This question has been discussed here. Check it out

http://forums.studentdoctor.net/showthread.php?t=297193

 

[faroo7a]

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

 

[recyrb]

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

 

[Slaughter421]

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.

 

[faroo7a]

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