Concept explainers
Interpretation:
The unbranched 1-alkene that will be liquid at room temperature has to be chosen from the given set of options.
Concept Introduction:
Organic compounds are represented shortly by the molecular formula and structural formula. Each and every compound has its own molecular formula. Compounds can have same molecular formula but not same structural formula.
Alkenes and cycloalkenes are hydrocarbons. They are nonpolar molecules. Water is a polar molecule. Therefore, alkenes and cycloalkenes do not get solubilized in water. In other words, alkenes and cycloalkenes are insoluble in water.
Regarding density, alkenes and cycloalkenes have density lower than water. When alkenes and cycloalkenes are mixed with water, two layers are formed which is a result of insolubility. Alkenes and cycloalkenes are present on top of water layer which is due to lesser density.
Boiling point of alkenes and cycloalkenes increase with an increase in carbon‑chain length or the ring size. The continuous chain alkenes which contain two to four carbon atoms are gases at room temperature. The continuous chain alkenes that contain five to seventeen carbon atoms and one double bond are liquids at room temperature.
When branching happens in the carbon chain, it lowers the boiling point of alkenes. In simple words, unbranched alkenes have more boiling point than branched alkenes with the same number of carbon atoms.
Cycloalkenes have more boiling point compared to noncyclic alkenes with the same number of carbon atoms. This is due to the more rigid and more symmetrical structures that occur in cyclic systems. Cyclopropene and cyclobutene are relatively unstable compound and gets converted into other hydrocarbons.
Want to see the full answer?
Check out a sample textbook solutionChapter 13 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- Show work with explanation needed. Don't give Ai generated solutionarrow_forward7. Calculate the following for a 1.50 M Ca(OH)2 solution. a. The concentration of hydroxide, [OH-] b. The concentration of hydronium, [H3O+] c. The pOH d. The pHarrow_forwardA first order reaction is 46.0% complete at the end of 59.0 minutes. What is the value of k? What is the half-life for this reaction? HOW DO WE GET THERE? The integrated rate law will be used to determine the value of k. In [A] [A]。 = = -kt What is the value of [A] [A]。 when the reaction is 46.0% complete?arrow_forward
- 3. Provide the missing compounds or reagents. 1. H,NNH КОН 4 EN MN. 1. HBUCK = 8 хно Panely prowseful kanti-chuprccant fad, winddively, can lead to the crading of deduc din-willed, tica, The that chemooices in redimi Грин. " like (for alongan Ridovi MN نيا . 2. Cl -BuO 1. NUH 2.A A -BuOK THE CF,00,H Ex 5)arrow_forward2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forward
- The following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forward
- The following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forwardSelect all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward3. Draw diagrams to represent the conjugation in these molecules. Draw two types of diagram: a. Show curly arrows linking at least two different ways of representing the molecule b. Indicate with dotted lines and partial charges (where necessary) the partial double bond (and charge) distribution H₂N* H₂N -NH2arrow_forward
- Organic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning