CHEM 301 Hour Exam III | Reactions and Reagents

35 Unit III reactions. Front: reagents and starting bond. Back: what changes, where groups go, and 3D result. Chemistry words such as ketone, carbonyl, ether, syn, and anti are explained in plain language on the cards.

HBr or HCl (alkene, C=C)

Adds H and Br or Cl across C=C, turning it into C–C. Br/Cl is a halogen. Markovnikov: the halogen goes on the C bonded to more other carbons. The H goes on the other C. No fixed same-side/opposite-side result. A positively charged C can form, so an H or carbon group may shift before the halogen attaches.

HBr + ROOR (alkene, C=C)

ROOR is a peroxide. Adds H and Br across C=C. Anti-Markovnikov: Br goes on the C bonded to fewer other carbons. H goes on the other C. No fixed same-side/opposite-side result. No carbon-group shift. This peroxide rule works for HBr, not HCl or HI.

H3O+ or dilute H2SO4/H2O (alkene, C=C)

Adds H and OH across C=C to make an alcohol (a C–OH group). Markovnikov: OH goes on the C bonded to more other carbons. No fixed same-side/opposite-side result. A positively charged C forms, so an H or carbon group may shift.

1) Hg(OAc)2, H2O 2) NaBH4 (alkene, C=C)

Adds H and OH across C=C to make an alcohol (C–OH). Markovnikov: OH goes on the C bonded to more other carbons. No dependable same-side/opposite-side rule for the final product. No carbon-group shift.

1) BH3·THF 2) H2O2, NaOH (alkene, C=C)

Adds H and OH across C=C to make an alcohol (C–OH). Anti-Markovnikov: OH goes on the C bonded to fewer other carbons. Syn: H and OH add from the same face of the double bond. No carbon-group shift.

H2 + Pt/Pd/Ni (alkene, C=C)

Adds one H to each C of the double bond, changing C=C to C–C. Syn: the two H atoms add from the same face. The product is an alkane (only single C–C bonds in that part).

Br2 or Cl2 (alkene, C=C)

Adds one Br or Cl to each C of the double bond; C=C becomes C–C. Anti: the two halogens add from opposite faces. No Markovnikov choice because both added atoms are the same. No carbon-group shift.

Br2/H2O or Cl2/H2O (alkene, C=C)

Adds OH and Br or Cl across C=C. OH goes on the C bonded to more other carbons; the halogen goes on the other C. Anti: they add from opposite faces. Product has an OH and a halogen on neighboring carbons.

Br2 or Cl2 in ROH (alkene, C=C)

R stands for a carbon-containing group. ROH means an alcohol, such as CH3OH. Adds OR and Br or Cl across C=C; OR means O bonded to the R carbon group. OR goes on the C bonded to more other carbons. Anti: OR and halogen add from opposite faces. Product contains an ether link (C–O–C).

1) RCO3H or mCPBA 2) H3O+ (alkene, C=C)

RCO3H and mCPBA are peroxy acids. First, O joins the two double-bond carbons in a 3-membered ring (an epoxide). Water/acid opens the ring: one OH ends up on each neighboring C. Overall anti: the OH groups are on opposite faces.

OsO4/NMO (alkene, C=C)

Adds an OH to each C of the double bond, making two OH groups on neighboring carbons. Syn: both OH groups add from the same face. The C=C becomes C–C.

1) O3 2) DMS or Zn/H2O (alkene, C=C)

O3 cuts the C=C bond. Each former double-bond C becomes C=O (a carbonyl). A C=O with an H on that C is an aldehyde; a C=O with carbon groups on both sides is a ketone. DMS or Zn/H2O is the workup that stops further oxidation.

1) O3 2) H2O2 (alkene, C=C)

O3 cuts the C=C bond. Each former double-bond C becomes C=O (a carbonyl). If that C had an H, H2O2 takes it further to a carboxylic acid (–COOH). If it had carbon groups on both sides, it stays a ketone (C=O between two carbons). A terminal =CH2 end can become CO2.

PBr3 (1° or 2° alcohol, C–OH)

Replaces OH with Br. The C–OH carbon keeps its other bonds. If that C has four different groups attached, its 3D arrangement flips (inversion). 1° means the OH-bearing C is bonded to one other C; 2° means two.

NaNH2 (terminal alkyne, C≡C–H)

Removes the H attached to the end of C≡C. This leaves a negatively charged end carbon, called an acetylide, which can bond to another carbon in a later step.

1) Excess NaNH2/NH3 2) H2O (two halogens on one C or neighboring C atoms)

Removes two HX pieces, where X is a halogen such as Br or Cl, to form a C≡C triple bond. H2O is the final workup.

1) Br2 2) Excess NaNH2 3) H2O (alkene, C=C)

Turns C=C into C≡C at the same two carbons. Br2 first adds one Br to each of the neighboring carbons. NaNH2 then removes two HBr pieces. H2O finishes the reaction.

1) NaNH2 2) methyl or 1° R–X (terminal alkyne, C≡C–H)

NaNH2 removes the end H from C≡C. The end carbon then bonds to the carbon bearing X in R–X (X is a halogen), making a new C–C bond. Best with methyl or 1° R–X; 1° means that X-bearing C has one other C attached.

Excess H2 + Pt/Pd/Ni (alkyne, C≡C)

Adds two H2 molecules across C≡C and changes it all the way to C–C. The result is an alkane. With this catalyst, the reaction does not usually stop at C=C.

H2/Lindlar catalyst (alkyne, C≡C)

Adds one H to each triple-bond C and stops at C=C. Syn: both H atoms add from the same face, giving a cis/Z alkene when each C has different groups (the carbon groups end up on the same side).

Na/liquid NH3 (internal alkyne, C≡C)

Adds one H to each triple-bond C and stops at C=C. Anti: H atoms add from opposite faces, giving a trans/E alkene when each C has different groups (the carbon groups end up on opposite sides).

1 equivalent HX (alkyne, C≡C)

One equivalent means one HX molecule per alkyne molecule. HX is an acid such as HBr or HCl. One H and one X add across C≡C, leaving C=C. Markovnikov: X goes on the C bonded to more other carbons. X is attached directly to a double-bond C. No fixed same-side/opposite-side result.

Excess HX (alkyne, C≡C)

HX is an acid such as HBr or HCl. Adds two H and two X across C≡C, changing it to C–C. Usually both X atoms end up on the same C, the one bonded to more other carbons (Markovnikov).

HgSO4, H2SO4, H2O (terminal alkyne, C≡C–H)

Adds H and OH across C≡C, with OH on the C bonded to more other carbons (Markovnikov). Bonds then shift to C=O (a carbonyl). For a terminal alkyne, the product is a methyl ketone: R–C(=O)–CH3. A ketone has C=O between two carbons.

1) Bulky R2BH 2) H2O2, NaOH (terminal alkyne, C≡C–H)

Adds H and OH across C≡C, with OH on the end C bonded to fewer other carbons (anti-Markovnikov). The initial addition is syn (same face). Bonds then shift to an aldehyde, R–CH2–CHO: an aldehyde is C=O with an H on the C=O carbon.

1 equivalent Br2 or Cl2 in CCl4 (alkyne, C≡C)

One equivalent means one Br2 or Cl2 molecule per alkyne molecule. Adds one halogen to each triple-bond C and stops at C=C. Anti: halogens add from opposite faces, usually giving an E/trans double bond with the halogens on opposite sides.

Excess Br2 or Cl2 in CCl4 (alkyne, C≡C)

Adds two halogen atoms to each of the two triple-bond carbons. C≡C becomes C–C, with four halogens total.

HBr (3° alcohol, C–OH)

Replaces OH with Br. 3° means the OH-bearing C is bonded to three other carbons. It proceeds through a positively charged C, so the 3D arrangement is generally mixed if that center can be stereogenic.

NaCN (1° alkyl halide, C–X)

CN replaces X (a halogen such as Br or Cl) and makes a new C–C bond. Product has C≡N, called a nitrile. 1° means the X-bearing C is bonded to one other C. If that C has four different groups attached, its 3D arrangement flips.

Br2, hν (alkane, only C–C single bonds)

hν means light. Replaces one C–H with C–Br. Br tends to replace H on the C bonded to the most other carbons. A chain reaction involving short-lived particles with one unpaired electron (radicals) carries it forward; more than one site can react if different C–H positions exist.

NBS, hν (alkene with a C–H next to C=C)

hν means light. Replaces an H on the carbon NEXT TO C=C with Br; the C=C is usually left in place. NBS is a bromine source. If the resulting short-lived particle with an unpaired electron can put the double bond in two positions, more than one product is possible.

NaOEt (alkyl halide, C–X)

NaOEt is sodium ethoxide, a small base. It often removes H from a C next to the C bearing X while X leaves, forming C=C (E2). Often favors the double bond whose carbons are bonded to more other carbons (Zaitsev). The H and X must line up on opposite sides in one plane. With a 1° C–X, substitution can compete: OEt replaces X, making an ether (C–O–C).

t-BuOK (alkyl halide, C–X)

t-BuOK is a bulky base. Removes H from a C next to the C bearing X while X leaves, forming C=C (E2). Often favors the double bond whose carbons are bonded to fewer other carbons (Hofmann). The H and X must line up on opposite sides in one plane.

Concentrated H2SO4, heat (alcohol, C–OH)

Removes OH and an H from a neighboring C as H2O, forming C=C. Often favors the double bond whose carbons are bonded to more other carbons. For 2°/3° alcohols a positively charged C can form, so an H or carbon group may shift; 2°/3° means the OH-bearing C has two/three C neighbors.

1) TsCl, pyridine 2) t-BuOK (alcohol, C–OH)

TsCl/pyridine turns OH into a better leaving group without changing its bond to C. Bulky t-BuOK then removes H from a neighboring C as that group leaves, making C=C (E2). Often favors the double bond with fewer carbon groups attached. H and leaving group must line up on opposite sides in one plane.