Determine a synthesis to prepare 2-chloro-4-methylpentane from 1-iodo-4-methylpentane.
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1-iodo-4-methylpentane is reacted with KOt-Bu(potassium tert-butoxide) to afford 4-methyl-1-pentene(E2 Reaction). Next, this product is reacted with HCl to perform a markovnikov addition addition across a double bond and afford the desired product, 2-chloro-4-methylpentane.
MendelSet practice problem # 1336 submitted by reifa on October 26, 2011.
Rank each set of compounds in order of decreasing boiling point (1 = highest boiling point):
a) ethane, n-octane, n-pentane
b) n-butane, 1-butanol, 1-chlorobutane.
c) n-octane, 2-methylheptane, 2,5-dimethylhexane
(Note that the n- prefix before an alkane just means that it's one chain, without any branching.)
Remember the rules of what gives compounds a higher boiling point (and melting point):
Van der Waals Forces
More electrons = more Van der Waals interactions = higher boiling point
So in general, the more "stuff" in a molecule (the higher the molecular weight), the higher its b.p. Example: C10H22 will have a higher b.p. than C5H12.
How to remember this trend? What do you grill with? propane gas. (C3H8)
What's in a lighter? butane liquid (C4H10)
Hydrogen Bonding
Compounds that can hydrogen bond have higher b.p./m.p/ than those that don't.
Compounds can hydrogen bond is they have a N, O, or F bonded directly to an H.
In organic chemistry, the best examples are alcohols (ROH) and amines (RNH2).
Branching
Branching makes it harder for molecules to pack together, which makes it harder to form Van der Waals interactions, and so tends to lower b.p./m.p.
a) (highest b.p.) n-octane > n-pentane > ethane (lowest b.p.)
Reason: octane has the most "stuff" (higher molecular weight).
b) 1-butanol > 1-chloroethane > n-butane
Reason: 1-butanol can hydrogen bond. 1-chloroethane has a higher molecular weight than n-butane.
c)n-octane > 2-methylheptane > 2,5-dimethylhexane
Reason: Branching. n-octane has no branching. 2,5-dimethylhexane has the most branching. Notice that each compound has the same molecular formua- C8H18. (and therefore the same weight).
MendelSet practice problem # 1287 submitted by Matt on October 2, 2011.
For the molecule shown below, indicate the hybridization (sp3, sp2, sp, etc.) of atoms A through H, and the bond angles of X, Y, and Z.
The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.
Also remember these rules:
If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp3 hybridized (s1 + p3 = 4).
If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp2 hybridized (s1 + p2 = 3).
If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s1 + p1 = 2).
Asp2 This oxygen has one double bond and two lone pairs. Three things total = sp2.
Bsp2 This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp2.
Csp3 This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp3.
Dsp3 This carbon has one implicit hydrogen and three single bonds. Four things total = sp3.
Esp3 This oxygen has two lone pairs and two bonding pairs. Four things total = sp3.
Fsp2 This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp2.
Gsp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.
Hsp This nitrogen has a lone pair and a triple bond. Two things total = sp.
For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:
Best way to separate two things is a straight line (180°).
Best way to separate three things is a "trifecta" (120°).
Best way to separate four things in three dimensional space is a tetrahedron (109.5°).
X 120° The carbon between the two oxygens is sp2 hybridized, so the shape is trigonal planar and the bond angle is 120°.
Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.
Z 109.5° The carbon is sp3 hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.
MendelSet practice problem # 1286 submitted by Matt on October 2, 2011.