12-14 reverse,

Question 1

halogenation of OH by HX
O grabs H from H-X, becomes -OH2+
second step X- replaces OH2+
Overall its an a transformation adding X to C

Answer 1

HX

X=Cl, Br, I

Question 2

Halogenation of alcohol

OH replaced by Cl

Answer 2

SOCl2

also requires mild base, poor nucleophile

Question 3

Halogenation of OH

OH replaced by Br

Answer 3

PBr3

requires mild base, poor nucleophile

Question 4

Sn2
LG leaves ( Cl, Br, I, OT)
nucleophile comes on, reagent replaces LG
switches enantiomers (R to S or vis)

Answer 4

SN2
Good Nucleophile ( strong base) Br, Cl, I

Question 5

LG out replaced by OH

enantiomers switch

Answer 5

Sn2

NaOH or KOH

Question 6

LG leaves replaced by OR

Answer 6

Sn2

NaOR (Na+ OR-)

Question 7

R-C (triple bond )C-H
replace H with another group
ex 
R-C triplebond C
to r-C TB C-CH2-R

Answer 7

1)strong base NaNH2
2) R-CH2-LG
LG = Cl, Br, I, Ts

Question 8

alkyl halide (r-X) or tosylate (R-OTs) to alkene
good leaving group comes off along with H (x and H)
double bond forms C=C
can form both E and Z

E1

Answer 8

weak base
ethanol (or polar protic solvent ( has Hs on N,O,X))
Heat

E1

Question 9

allylic or benzyllic OH to alkene
OH is secondary or tertiary
OH and H come off
C=C forms between two C’s

Answer 9

acid (H2SO4 or H3PO4)

Heat

Question 10

1 step no carbo cation
methyl leaving group or no carbocations or beta-H’s
has to be methyl, primary, secondary or 1* w/ allylic/benzyllic LG
not tertiary
Good nuclei-strong base-cool
or Good nuk-weak-aprotic

Answer 10

SN2

Question 11

2 Steps carbo forms,
Only secondary or Tert or 1* if allylic/benzyllic LG
Good nuke-weak base- polar protic ( h on NOX)
or
Poor nuk-weak base or polar protic solvent-cool

Answer 11

Sn1

Question 12

1 Step no carbo forms
1, 2, 3*
strong base, poor nuk, hot

Answer 12

E2

Question 13

2 Step, carbo forms
1* if allylic or benzyylic LG, 2, 3
WEAK Base only, usually hot,

poor nun, weak base (protic solvent), Hot

Answer 13

E1