
(a)
Interpretation: Structural formula for 1-Pentene is to be drawn.
Concept Introduction:
(a)

Answer to Problem 23LC
CH3-CH2-CH2-CH=CH2
Explanation of Solution
The compound contains 5 carbon atoms as it has “Pent” in the name. For “ene” the compound contains a double bond. For “1-ene” the compound contains a double bond on the 1-C atom.
Structural Formula is
CH3-CH2-CH2-CH=CH2
b.
Interpretation: Structural formula for 2-hexene is to be drawn.
Concept Introduction: Alkenes are comprised of carbon and hydrogen with at least one double bond in the carbon chain.
b.

Answer to Problem 23LC
CH3-CH2-CH2-CH=CH-CH3
Explanation of Solution
The compound contains 6 carbon atoms as it has “Hex” in the name. For “ene” the compound contains a double bond. For “2-ene” the compound contains a double bond on the 2-C atom.
Structural Formula is
CH3-CH2-CH2-CH=CH-CH3
c.
Interpretation: Structural formula for 2-methyl-1-butene is to be drawn.
Concept Introduction: Alkenes are comprised of carbon and hydrogen with at least one double bond in the carbon chain.
c.

Answer to Problem 23LC
Explanation of Solution
The compound contains 4 carbon atoms. For “ene” the compound contains a double bond. For “2-ene” the compound contains a double bond on the 1-C atom. For “2-methyl” the compound contains a methyl group on the 2-C atom.
Structural Formula is
d.
Interpretation: Structural formula for 2,5-dimethyl-3-hexene is to be drawn.
Concept Introduction: Alkenes are comprised of carbon and hydrogen with at least one double bond in the carbon chain.
d.

Answer to Problem 23LC
Trans-2,5-dimethyl-3-hexene Cis-2,5-dimethyl-3-hexene
Explanation of Solution
The compound contains 6 carbon atoms as it has “hex” in the name. For “ene” the compound contains a double bond. For “3-ene” the compound contains a double bond on the 3-C atom. For “2,5-dimethyl” the compound contains methyl groups on the 2-C atom & 5-C atom.
It has both Trans-2,5-dimethyl-3-hexene Cis-2,5-dimethyl-3-hexene Structural Formula is
Trans-2,5-dimethyl-3-hexene Cis-2,5-dimethyl-3-hexene
Chapter 22 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- What would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forwardPredict the products of this organic reaction: Explanation Check IN NaBH3CN H+ ? Click and drag to start drawing a structure. D 5 C +arrow_forwardPredict the products of this organic reaction: H3O+ + ? • Draw all the reasonable products in the drawing area below. If there are no products, because no reaction will occur, check the box under the drawing area. • Include both major and minor products, if some of the products will be more common than others. • Be sure to use wedge and dash bonds if you need to distinguish between enantiomers. No reaction. Click and drag to start drawing a structure. dmarrow_forward
- Iarrow_forwardDraw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward
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