• Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

Card Range To Study

through

image

Play button

image

Play button

image

Progress

1/18

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

18 Cards in this Set

  • Front
  • Back
Small / Negative pKa
More acidic
Less basic
Higher number pKa
Less acidic
Strong base
Add alcohol group to alkene
Alkene + H2O and H2SO4
Halogenation of alkene
X-cl, X-Br + Alkane
Halogenation, either side of double bond.
Br-Br, cl-cl + Alkene (Trans, if cyclic)
Anti Markonicov
Diboron hexahydride + h2o2
Form 2 alcohol groups (diol) from alkene.
Osmium tetroxide oso4 (cis)
Catalytic reduction of alkene
Hydrogren (h2) + Platinum or paladium. Forms alkane.
State Markovnikovs rule
When halogenating an alkene, the hydrogen will attach to the most hydrogenated carbon.
Cloronation of benzene
Cl2 + AlCl3
Bromination of benzene
Br2 + FeBr3
Oxidation of benzilic position
Turns carbon at methyl position into carboxcyllic acid cooh.
Chromic acid
potassium dichromate
potassium permangenate
Nitration of benzene
Benzene + HNO3 AND H2SO4.
Produces benzene with NO2 attached.
Sulfination of benzene
Benzene + H2SO4
Benzene with SO2OH attached
Benzene sulfonic acid
Alkylation (Adding methyl group to benzene)
X-cl + AlCl3 to benzene, creates toluene (methylbenzene)
Acylation (Double bonded O + cl)
Benzene - bonded to C - double bonded to o AND whatever akyl group was attached
State Zaitsevs rule
The major product of an elimination reaction involving alkenes will be the most stubstituted product.
Form alkene from alkane
Strong base (Ethoixide - OCH2CH3)
Removes substituent - Br or Cl